Finding the graph formula with know points and other equation.

In summary, the conversation discusses an equation for determining the acceleration of an object with a constant power source and air resistance, as well as trying to graph it as a velocity-time graph. The conversation also mentions a similar equation for a constant friction force and the maximum speed in this scenario. The conversation then delves into using differential equations to solve for the velocity, but the equations are not formatting correctly. The conversation concludes with a request for advice on how to incorporate time into the equation without velocity on both sides.
  • #1
Pharrahnox
106
0
I have an equation for determining the acceleration of an object being propelled by a constant power source, that is affected by air resistance:

a = [itex]\frac{P}{mv}[/itex]-[itex]\frac{CDpAv2}{2m}[/itex]

Since F = [itex]\frac{P}{v}[/itex]

I am trying to graph this as a velocity-time graph, however, I don't know how to do it. There is no time variable that I can replace with x, and the y-value (velocity) is mixed into the equation already.

I remember the equation given to me for a similar sort of thing, without air resitance, but instead just a constant friction force, that was something like this:

y = k(1-e-ax)

Where k is a constant, which is the maximum speed, and a is another constant which represents the force of air resistance.

The maximum speed in this case is [itex]\sqrt[3]{\frac{2P}{CDpA}}[/itex], so the equation would be something like:

y = [itex]\sqrt[3]{\frac{2P}{CDpA}}[/itex](1-e-ax)

But that's as far as I've gotten. By the use of iteration, I have determined the velocity at several different times, here's a few, just in case it helps:

(0,0) (25,83.4762) (50,118.1195) (75,126.1601) (100,127.7624) (125,128.0709)

This is for variables of values: p = 0.001, A = 1900.933, m = 1900.933, P = 1*106
and CD = 0.5

Any help would be greatly appreciated, and if you need any more information, just let me know.EDIT: the equations don't seem to be formatting correctly, so I'll redo them down here:

a = P/v - (Cd*p*A*v^2) /2
F = P/v
max speed = ( (2*P) / (Cd*p*A) )^1/3
y = ( (2*P) / (Cd*p*A) )^1/3 * (1 - e^-ax)
 
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  • #2
Given ##a=\frac{dv}{dt}##, what you have is a separable ordinary differential equation [tex]\frac{dv}{dt}=\frac{\alpha}{v}-\beta v^2[/tex] where ##\alpha=\frac{P}{m}## and ##\beta=\frac{C_DpA}{2m}## have been substituted to make the equation (slightly) more manageable. Hopefully that's enough to get you going.

I'm not sure where you got your force equation. Are you not using ##F=ma##? Furthermore, I don't know that it's even useful here.

I reckon it's doubtful that the equation that works in a constant friction situation applies here. So you're likely barking up the wrong tree there.
 
  • #3
I have never done differential equations before, and I tried to just find the anti-derivative, but my teacher said that I had done it incorrectly, as I differentiated dv instead of dt, or the other way around. Anyway, here is what I got:

v = (Pt/m)*ln(|v|)-(CDpAv3t)/(6m)+c

Unfortunately from there, if it is correct, I don't know where to go - how to get v by itself without v on the other side.

What do I do from here, or from the start if that isn't correct?

Thanks for your response.
 
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  • #4
Pharrahnox said:
the equations don't seem to be formatting correctly

I think this is due to your using the SUB and SUP tags instead of the usual LaTex notation.

[itex] a = \frac{P}{v} - \frac{ (C_d p A v^2) }{2} [/itex]
 
  • #5
Oh ok, I'll give it a go:

a = ([itex]\frac{Pt}{m}[/itex])ln(|v|)-[itex]\frac{C_{D}pAv^{3}t}{6m}[/itex]

Seems to work, thanks.
 
  • #6
Does anyone have any advice on what I should do to get time into the equation, without velocity on both sides? I don't even know what to type into google to find information on it. Any information would be much appreciated.
 

FAQ: Finding the graph formula with know points and other equation.

What is a graph formula?

A graph formula is a mathematical expression that describes the relationship between two variables on a graph. It can be used to plot points on a graph and to predict the value of one variable based on the value of the other variable.

How do I find the graph formula with known points?

To find the graph formula with known points, you can use the slope-intercept form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept. You can plug in the coordinates of the known points and solve for the values of m and b.

Can I find the graph formula if I only have one known point?

No, you need at least two known points to find the graph formula. With one point, you can only determine a single value on the graph, but not the entire formula.

Can I use any equation to find the graph formula?

No, the equation must represent a linear relationship between the two variables. This means that the variables must be raised to the first power and there cannot be any exponents or variables in the denominator.

Can I use the graph formula to predict values outside of the known points?

Yes, the graph formula can be used to predict values outside of the known points. This is one of the main purposes of finding the graph formula - to be able to make predictions and analyze the relationship between the variables on the graph.

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